SOLUTION: find all other zeros give 2i is one zero. f(x) = x^5 - 3x^4 + 5x^3 - 15x^2 + 4x - 12

Algebra ->  Equations -> SOLUTION: find all other zeros give 2i is one zero. f(x) = x^5 - 3x^4 + 5x^3 - 15x^2 + 4x - 12      Log On


   



Question 1010315: find all other zeros give 2i is one zero.
f(x) = x^5 - 3x^4 + 5x^3 - 15x^2 + 4x - 12

Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39625) About Me  (Show Source):
You can put this solution on YOUR website!
The TWO imaginary zeros are 2i and -2i, giving the quadratic factor %28x-2i%29%28x-%28-2i%29%29=%28x-2i%29%28x%2B2i%29=highlight_green%28x%5E2%2B4%29.

Use polynomial division to obtain the degree three factor of the function, and use whatever you know to further factorize it, such as try Rational Roots Theorem and synthetic division to check possible roots.

The result from the polynomial division (process not shown here) is
f%28x%29=%28x%5E2%2B4%29%28x%5E3-3x%5E2%2Bx-3%29.

You now want to look for up to three possible real, rational roots among the possibilities of -1, -3, 1, 3. The only one of those that worked was....
3, for the linear binomial factor x-3 and quotient x%5E2%2B1.

The listing of these three other roots then is:
3, i, -i.

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!

find all other zeros give 2i is one zero.
f(x) = x^5 - 3x^4 + 5x^3 - 15x^2 + 4x - 12
One zero is 2i, so the other, its conjugate is: - 2i
This gives us: (x - 2i)(x + 2i), or x%5E2+%2B+4
Dividing x%5E5+-+3x%5E4+%2B+5x%5E3+-+15x%5E2+%2B+4x+-+12 by x%5E2+%2B+4 results in:
x%5E3+-+3x%5E2+%2B+x+-+3, which can be factored into:
x%5E2%28x+-+3%29+%2B+1%28x+-+3%29
%28x%5E2+%2B+1%29%28x+-+3%29
%28x+%2B+i%29%28x+-+i%29%28x+-+3%29.
This leads to zeroes of: - i, i, and 3.
Thus the zeroes of f%28x%29+=+x%5E5+-+3x%5E4+%2B+5x%5E3+-+15x%5E2+%2B+4x+-+12 are: highlight_green%28system%28x+=%22+%22%2B-+2i%2C+x+=%22+%22+%2B-i%2C+x+=+3%29%29